I am based in the answers given for my question in this post.
Let $ \, S \, $ be a semigroup such that $ \ |S| \geq 2 \ $. Its power semigroup is the power set $ \, \wp(S) \, $ together with the binary operation $$XY = \{ xy \in S : x \in X, \ y \in Y \} \ \ .$$
I am interested in the semigroup $ \ Q_S = \wp(S) \setminus \{ \varnothing \} \ $, which I also call the power semigroup of $ \, S \, $.
A semigroup $ \, S \, $ is said left cancellative if, and only if, for all $ \ x,y,z \in S \ $, if $ \ xy=xz \ $, then $ \ y=z \ $.
A semigroup $ \, S \, $ is said right cancellative if, and only if, for all $ \ x,y,z \in S \ $, if $ \ yx=zx \ $, then $ \ y=z \ $.
As the answers in the linked post proves that there is no cancellative power semigroup for such $ \, S \, $ I wonder if there is at least a left cancellative one.
I would like to see an example of a semigroup $ \, S \, $ such that $ \ |S| \geq 2 \ $ and $ \, Q_S \, $ is left cancellative.
Of course, if there is such an $S$, then $ \, Q_S \, $ will obviously NOT be right cancellative.
Take any right zero semigroup $S$ (it satisfies $xy=y$). Then its power semigroup is right zero and satisfies left cancelation.
The converse statement that if the power semigroup has left cancelation then the semigroup satisfies $xy=y$ is also true. Indeed, suppose $Q_S$ has left cancelation. Then for every $s\in S$ you have $\{s^n, n\ge 1\}\{s\}=\{s^n, n\ge 1\}\{s^n, n\ge 1\}$, so $\{s^n, n\ge 1\}=\{s\}$. In particular, $s=s^2$ for every $s$. Then $\{s\}\{st\}=\{s\}\{t\}$ for every $t\in S$. Hence $st=t$ for every $s, t\in S$ and $S$ consists of right zeroes.